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Question
Three vectors A, B and C add up to zero. Find which is false.
Options
(A × B) × C is not zero unless B, C are parallel
(A × B) . C is not zero unless B, C are parallel
If A, B, C define a plane, (A × B) × C is in that plane
(A × B) . C = |A| |B| |C| → C2 = A2 + B2
Solution
If A, B, C define a plane, (A × B) × C is in that plane
Explanation:
These types of problems can be solved by hit and trial method by picking up options one by one
Sum of vectors is given as `vecA + vecB + vecC` = 0
Hence, we can say that `vecA, vecB` and `vecC` are in one plane and are represented by the three sides of a triangle taken in one order. Let us discuss all the options one by one.
a. We can write
`vecB xx (vecA + vecB + vecC) = vecB xx 0 = 0`
⇒ `vecB xx vecA + vecB xx vecB + vecB xx vecC = 0`
⇒ `vecB xx vecA = - vecB xx vecC`
⇒ `vecA xx vecB = vecB xx vecC`
∴ `(vecA xx vecB) xx vecC = (vecB xx vecC) xx vecC`
It cannot be zero.
If `vecB || vecC`, then `vecB xx vecC` = 0, then `(vecB xx vecC) xx vecC` = 0.
Even if `vecA || vecC` or `vecA || vecB`, then also the cross product in any manner will be zero. So option (a) is not false.
b. `(vecA xx vecB). vecC = (vecB xx vecA) . vecC` = 0 whatever be the positions of `vecA, vecB` and `vecC`. If `vecB || vecC`, then `vecB xx vecC` = 0, then `(vecB xx vecC)` = 0. So, option (b) is not false.
c. `(vecA xx vecB) = vecX = vecAvecB sin theta` and the direction of X is perpendicular to the plane containing `vecA` and `vecB. (vecA xx vecB) xx vecC = X xx vecC`. Its direction is in the plane of `vecA, vecB` and `vecC`. So, option (c) is false.
d. If `vecC^2 = vecA^2 + vecB^2`, then angle between `vecA` and `vecB` is 90°.
∴ `(vecA xx vecB) * vecC = (vecAvecB sin 90° X) . vecC = vecAvecB (X * vecC)` = ABC cos 90° = 0. Hence, option (d) is not false.
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